Intersection with Complement is Empty iff Subset

Theorem

$S \subseteq T \iff S \cap \map \complement T = \O$

where:

$S \subseteq T$ denotes that $S$ is a subset of $T$

$S \cap T$ denotes the intersection of $S$ and $T$

$\O$ denotes the empty set

$\complement$ denotes set complement.

$S \cap T = \O \iff S \subseteq \relcomp {} T$

$\ds S$

$\subseteq$

$\ds T$

$\ds \leadstoandfrom \ \ $

$\ds S \setminus T$

$=$

$\ds \O$

$\ds \leadstoandfrom \ \ $

$\ds S \cap \map \complement T$

$=$

$\ds \O$

$\blacksquare$

Some sources present this as an alternative definition of a subset:

Set $A$ is a subset of set $B$ if set $A$ contains no member that is not also in set $B$

but this is not mainstream.

1955: John L. Kelley: General Topology ... (previous) ... (next): Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Exercise $3.3 \ \text{(b)}$
1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{B vi}$
1971: Patrick J. Murphy and Albert F. Kempf: The New Mathematics Made Simple (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets: Subsets: Definition: $1.7$